A New Number and Logic Puzzle

New INTERACTIVE Version

59 DeGarmo Hills Road

Wappingers Falls, NY 12590

SumSum

SumSum

At first the SumSum

When you complete the grid there will be two black boxes in each line, that is, in each horizontal row and each vertical column. (Three boxes will be black in puzzles with more than 10 digits.) These two black boxes will never touch each other. The two black boxes divide the line into 1, 2 or 3 parts, called runs. If both black boxes are at the edges of the grid, then there is only one run. If only one of the black boxes is at an edge, then there are two runs. If neither of the black boxes is at an edge of the grid, then they separate that line into three runs.

28 |

5,23 |

5,10,13 |

The remaining boxes in each line are white. As you solve the puzzle, you will write one digit into each of the white boxes. Each digit will appear only once in any line.

The numbers to the left of each row and above each column are the sums. Each sum is the total of the digits in each run in that line. The sums are given in order, left to right, or top to bottom. For example, if there are three sums shown to the left of a row, then the first sum is the total of the digits in the first run, the second sum is the total of the digits in the second run, and the third sum is the total of the digits in the third run. For example,

7,3,18 | 4 | 3 | 2 | 1 | 7 | 5 | 6 |

In this example the first run is 4+3. That totals to 7. The second run is 2+1, which totals 3. The third run is 7+5+6, which totals 18. The totals of these three runs match the three sums at the left, 7, 3 and 18 in order.

The best way to solve a SumSum puzzle is to start at the edges and work inward. At the start you don't know which boxes are black and which are white, so they will be shown as gray. Your first step is to identify which boxes must be black, and which must be white. Gradually you will turn the puzzle from all gray to a pattern of black and white. If you are solving the puzzle interactively on this website, you simply click on a box, and then click either Black or White.

Some black boxes can be placed immediately. For example, if a line has only one sum given, then the two end boxes in that line must both be black, like this

28 |

The rest of the boxes will contain numbers, so they will be white, like this

28 |

Another case where you can place a black box immediately is when there are three sums given for a line, and either the first or last sum is 1 or 2. In this case, the end box must contain a 1 or 2, and the next box must be black. For example,

1,8,19 | 1 | ||||||||

12,14,2 | 2 |

It is also valuable to mark boxes that cannot be black. When a row or column has three sums, the two boxes at the ends of that line cannot be black. Often more than one box at the end of a line must be white. Suppose you are solving a 7-digit SumSum

23,1,4 |

There are only 4 boxes left. Two of them must be black, and the other two must be 1 and 4, like this

23,1,4 | 1 | 4 |

When you figure out how many white boxes are needed you should take into account any digits that have already been placed. A sum of 13 could be achieved with just 2 white boxes, namely 6+7, but in the next example a 3 and a 1 have already been placed, so 4 white boxes are needed. Three white boxes are not enough because 3+7+1 is only 11. So

13,7,8 | 3 | 1 |

13,7,8 | 3 | 1 |

Since the sum for the rightmost run is 8, which requires at least 2 white boxes, the only possible placement of the black and white boxes is

13,7,8 | 3 | 1 |

If you are solving SumSum puzzles using paper and pencil you might mark boxes that must be white with a . or - sign, as below.

15,3,10 | - | - | - | - | - |

If you are solving interactively on this website, undecided boxes will be colored gray, and boxes that you decide must be white will be colored white, like this

15,3,10 |

The pattern of black and white boxes can be completed by working the rows and columns simultaneously. As you mark boxes in the columns white, this reduces the number of boxes in the rows that could possibly be black, and vice-versa.

Let's look at some more situations where you can deduce whether a box must be black or white. One of the most obvious is when all of the boxes except two have been marked white. Those last 2 boxes must be black. So

10,8,10 |

10,8,10 |

Once you have marked two boxes in a line black all of the rest must be white. So

1,25,2 | 1 | 2 |

1,25,2 | 1 | 2 |

If there are 3 undecided boxes, and they are all adjacent, then the two outside boxes must be black, and the middle box must be white. So

11,6,11 |

11,6,11 |

If there are 3 undecided boxes, and they are in a 1 and 2 arrangement, or a 2 and 1 arrangement, you can immediately mark the singleton box black. So

10,10,8 |

10,10,8 |

If there are 3 undecided boxes, and they are all singletons, but the sum for the middle run is too large for the numbers to fit between any two of the undecided boxes, then the two outside boxes must be black and all of the remaining boxes must be white. In the next example, the middle sum is 8, which needs at least 2 white boxes. So

10,8,10 |

10,8,10 |

Once any box has been marked black, all of its immediate neighbors can be marked white. Thus

5,17,6 | |||||||||

5,13,10 | |||||||||

6,17,5 |

5,17,6 | |||||||||

5,13,10 | |||||||||

6,17,5 |

It is also valuable to know the maximum length for a run. A run with sum 1 or 2 can have only box. A run with sum 3 to 5 can have at most 2 boxes, because 3 boxes would total at least 1+2+3. A run with sum 6 to 9 can have at most 3 boxes, and so forth. Therefore the previous example becomes

5,17,6 | |||||||||

5,13,10 | |||||||||

6,17,5 |

Once a black box has been placed on a line, if you can determine which run that box belongs to, then you can mark additional white boxes that belong to the next run in the line, either to the right or the left. Continuing the previous example, on the top row the black box belongs to the first run whose sum is 5. The next run has a sum of 17, so the next 3 boxes must be white. On the third row the black box belongs to the last run, so the 3 boxes to its left must also be white. Therefore the previous example becomes

5,17,6 | |||||||||

5,13,10 | |||||||||

6,17,5 |

It is now clear that the one remaining undecided box on the top row must be black.

5,17,6 | |||||||||

5,13,10 | |||||||||

6,17,5 |

You can identify the most white boxes by looking at lines with 3 sums, but you can also identify some white boxes by looking at lines with 2 sums. Consider this row

20,8 |

One end box must be black, but you don't know which end. If the black box is at the left, then the boxes marked X in the top row must be white. If the black box is at the right, then the boxes marked X in the bottom row must be white.

20,8 | X | X | X | X | X | X | |||

20,8 | X | X | X | X | X | X |

Where the X's overlap, those boxes must be white in either case. You can mark those boxes white, like this.

20,8 |

In some cases it is possible to mark some of the center boxes white. Especially in larger puzzles, if the sums for the outer runs are both small much of the center of a line can be marked white. Look at this example.

3,21,4 |

The outer runs have sums of 3 and 4. Either of these runs could be 1 box or 2 boxes. There are 3 possible arrangements, since the outer runs cannot both have 2 boxes. As before the boxes that must be white are marked X.

3,21,4 | X | X | X | X | X | X | X | ||

3,21,4 | X | X | X | X | X | X | X | ||

3,21,4 | X | X | X | X | X | X | X |

Every box that is marked X in all 3 of these arrangements can be marked white, like this

3,21,4 |

Once most of the black boxes have been placed it becomes time to fill the boxes with numbers. The simplest case is when a run has only one box. In this case the number is the same as the run sum. It can be filled in immediately.

5,13,10 | 5 | ||||||||

10,3,15 | 3 | ||||||||

8,16,4 | 4 |

Another simple case is when there is only one unfilled box in a run. In this case you just subtract the numbers in the boxes from the run sum, and the result is the missing number. Here the 3 numbers in the middle run must add to the sum 9. The numbers 2 and 3 have already been found. You subtract 2 and 3 from 9, like this 9-2-3, and the result is 4. That's the missing number that goes into the third box. Now the 3 numbers in the run, 2+4+3 add up to 9. So

8,9,11 | 2 | 3 |

8,9,11 | 2 | 4 | 3 |

When you have filled in all of the singletons it is time to work on runs where 2 or 3 numbers are missing. It is easiest to start with runs where the sum is close to the minimum or maximum value for a run of that length. For example, in a 7-digit SumSum a run of 3 numbers can range from 1+2+3, or 6, to as high as 7+6+5, or 18. The range is 6 to 18.

Runs with sums near these limits have the fewest possibilities. A sum of 7, near the bottom of the range, has only one possibity, namely 1+2+4. A sum of 16, near the top of the range, has two possibilities, either 3+6+7 or 4+5+7. By contrast a run of 12, which is right at the center of the range 6 to 18 has 5 possibilities, namely 1+4+7, 1+5+6, 2+3+7, 2+4+6 and 3+4+5. Since 3 numbers can be arranged in 6 different orders you would have 30 different arrangements of numbers for a run of 3 numbers that add to 12.

Use the other numbers in the same row, and numbers in the same columns to decide which of the possible arrangements fits. In the following example there are 4 ways to fill in the first run with a sum of 5, namely 14, 23, 32 or 41. Since there is already a 1 in the row that eliminates 14 and 41, so it narrows the possibities to 23 or 32. Since there is already a 3 in column 1, it cannot be 32. That leaves only 23. So

5,22,1 | 1 | ||||||||

3,25 | 3 |

5,22,1 | 2 | 3 | 1 | ||||||

3,25 | 3 |

Next is a variation on the same idea. There are 6 ways to get the sum of 9, namely 27, 36, 45, 54, 63 and 72. The sum of 7 can be obtained by any of the 6 arrangements of 1+2+4, namely 124, 142, 214, 241, 412 and 421. however, you know for certain that the third run will contain 1, 2 and 4. This eliminates 27, 45, 54 and 72, which narrows the possibities to 36 or 63. Since there is already a 3 in column 1, it cannot be 36. That leaves only 63. So

9,12,7 | |||||||||

3,25 | 3 |

9,12,7 | 6 | 3 | |||||||

3,25 | 3 |

Another way to reduce the possibilities is by comparing the row runs with the column runs. In the next example the sum for the run in column 1 is 11. The possibilities are 47, 56, 65 and 74. The possibilities for the first run on the top row are 14, 23, 32 and 41. The only number that occurs in the first position in both lists is 4. So the horizontal run must be 41 and the vertical run must be 47.

5,10,13 | |||||||||

10,10,8 |

5,10,13 | 4 | 1 | |||||||

10,10,8 | 7 |

5,10,13 | 4 | 1 | |||||||

10,10,8 | 7 | 2 | 1 |

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